a, \(x^2\)  - 19 = 5.9

     \(x^2\) - 19 = 45

     \(x^2\)         = 45 + 19

     \(x^2\)         = 64

      \(x^2\)        = 8²

      \(x\)         = 8 

b, (2\(x\) + 1)³ = -0,001

    (2\(x\) + 1)³ = (-0,1)³

     2\(x\) + 1   = -0,1

     2\(x\)        = -0,1 - 1

     2\(x\)       = - 1,1

       \(x\)      = -1,1: 2

       \(x\)      = -  0,55

\(x^2-19=5\cdot9\\\Rightarrow x^2-19=45\\\Rightarrow x^2=45+19\\\Rightarrow x^2=64\\\Rightarrow x^2=(\pm8)^2\)

\(\Rightarrow\left[{}\begin{matrix}x=8\\x=-8\end{matrix}\right.\)

\(---\)

\((2x+1)^3=-0,001\\\Rightarrow (2x+1)^3=(-0,1)^3\\\Rightarrow2x+1=-0,1\\\Rightarrow2x=-0,1-1\\\Rightarrow2x=-1,1\\\Rightarrow x=-1,1:2\\\Rightarrow x=\dfrac{-11}{20}\\---\)

\(\bigg(\dfrac56\bigg)^{2x-1}=\bigg(\dfrac56\bigg)^5\\\Rightarrow 2x-1=5\\\Rightarrow2x=5+1\\\Rightarrow2x=6\\\Rightarrow x=6:2\\\Rightarrow x=3\\---\)

\(\bigg(\dfrac13x-\dfrac23\bigg)^3=27\\\Rightarrow\bigg(\dfrac13x-\dfrac23\bigg)^3=3^3\\\Rightarrow\dfrac13x-\dfrac23=3\\\Rightarrow\dfrac13x=3+\dfrac23\\\Rightarrow\dfrac13x=\dfrac{11}{3}\\\Rightarrow x=\dfrac{11}{3}:\dfrac13\\\Rightarrow x=11\\---\)

\(\bigg(\dfrac{1}{32}\bigg)^x=\bigg(\dfrac12\bigg)^{15}\\\Rightarrow\bigg(\dfrac{1}{32}\bigg)^x=\bigg[\bigg(\dfrac{1}{2}\bigg)^5\bigg]^3\\\Rightarrow\bigg(\dfrac{1}{32}\bigg)^x=\bigg(\dfrac{1^5}{2^5}\bigg)^3\\\Rightarrow\bigg(\dfrac{1}{32}\bigg)^x=\bigg(\dfrac{1}{32}\bigg)^3\\\Rightarrow x=3\\Toru\)

3: \(\left|x-\dfrac{3}{4}\right|-\dfrac{1}{2}=0\)

\(\Leftrightarrow\left|x-\dfrac{3}{4}\right|=\dfrac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{1}{2}\\x-\dfrac{3}{4}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=\dfrac{1}{4}\end{matrix}\right.\)

Câu 1: 

=>15(2x+1)-8(3x-1)=100

=>30x+15-24x+8=100

=>6x+23=100

hay x=77/6

Câu 2:

=>2(5x-3)+12-3(7x-1)=x+2

=>10x-6+12-21x+3-x-2=0

=>-12x=-7

hay x=7/12

Câu 3: 

\(\Leftrightarrow2\left(x^2-1\right)+3\left(x+1\right)=2\left(x^2-4x+4\right)\)

\(\Leftrightarrow2x^2-2+3x+3-2x^2+8x-8=0\)

=>11x-7=0

hay x=-7/11

Câu 4:

(x - 4)^3/6 + 1 = x(x + 1)/2 - (x - 5)(x + 5)/3

<=> (x - 4)^3 + 6/6 = x^2 + x/2 - x^2 - 25/3

<=> (x - 4)^3 + 6/6 = 3x^2 + 3x - 2x^2 + 50/6

<=> (x - 4)^3 + 6 = 3x^2 + 3x - 2x^2 + 50

<=> x^3 - 12x^2 + 48x - 58 = x^2 + 3x + 50

<=> x^3 -13x^2 + 45x - 108 = 0

Đến đây bạn bấm máy nhẩm nghiệm là ra nhé

Câu 5:

3(x + 2)^3/5 - (x - 1)^2/10 = (x - 3)(x + 3)/2

<=> 6(x + 2)^3 - (x - 1)^2/10 = 5(x^2 - 9)/10

<=> 6(x + 2)^3 - (x - 1)^2 = 5(x^2 - 9)

<=> 6x^3 + 36x^2 + 72x + 48 - x^2 + 2x - 1 - 5x^2 + 45 = 0

<=> 6x^3 + 30x^2 + 74x + 92 = 0

Đến đây bạn bấm máy nhẩm nghiệm như câu 4 nhé

1) \(\left(x-2\right)\left(3+2x\right)-2x\left(x+5\right)=6\)

\(3x+2x^2-6-4x-2x^2-10x-6=0\)

\(-11x=12\)

\(x=-\dfrac{12}{11}\)

2) \(x^2-4-\left(x-5\right)\left(x-2\right)=0\)

\(\left(x-2\right)\left(x+2\right)-\left(x-5\right)\left(x-2\right)=0\)

\(\left(x-2\right)\left(x+2-x+5\right)=0\)

\(7\left(x-2\right)=0\)

\(\Leftrightarrow x=2\)

1, \(3x+2x^2-6-4x-2x^2-10x=0\Leftrightarrow-11x-6=0\Leftrightarrow x=-\dfrac{6}{11}\)

2, \(\left(x-2\right)\left(x+2\right)-\left(x-5\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+2-x+5\right)=0\Leftrightarrow x=2\)

3, bạn xem lại đề 

5, đk x khác -4 ; 4 

\(96=\left(2x-1\right)\left(x-4\right)+\left(3x-1\right)\left(x+4\right)-6\left(x^2-16\right)\)

\(\Leftrightarrow96=2x^2-9x+4+3x^2+11x-4-6x^2+96\)

\(\Leftrightarrow-x^2+2x=0\Leftrightarrow-x\left(x-2\right)=0\Leftrightarrow x=0;x=2\)(tm) 

3) 

\(\dfrac{x-3}{3}-\dfrac{x+2}{2}=\dfrac{x}{6}\\ \Leftrightarrow\dfrac{2\left(x-3\right)}{6}-\dfrac{3\left(x+2\right)}{6}=\dfrac{x}{6}\\ \Leftrightarrow2x-6-3x-6=x\\ \Leftrightarrow2x-3x-x=6+6\\ \Leftrightarrow-2x=12\\ \Leftrightarrow x=-6\)

Vậy PT có tập nghiệm S = { -6 }

\(Đặt:\left\{{}\begin{matrix}a=\dfrac{1}{\left|x+3\right|}\left(ĐK:x\ne-3\right)\\b=\dfrac{1}{\left|y\right|-2}\left(ĐK:y\ne\pm2\right)\end{matrix}\right.\\ Có:\left\{{}\begin{matrix}\dfrac{1}{\left|x+3\right|}+\dfrac{4}{\left|y\right|-2}=\dfrac{11}{6}\\\dfrac{5}{\left|x+3\right|}+\dfrac{2}{\left|y\right|-2}=\dfrac{11}{6}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a+4b=\dfrac{11}{6}\\5a+2b=\dfrac{11}{6}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a+4b=\dfrac{11}{6}\\10a+4b=\dfrac{22}{6}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-9a=-\dfrac{11}{6}\\a+4b=\dfrac{11}{6}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{11}{54}\\b=\dfrac{\dfrac{11}{6}-\dfrac{11}{54}}{4}=\dfrac{11}{27}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{\left|x+3\right|}=a=\dfrac{11}{54}\\\dfrac{1}{\left|y\right|-2}=b=\dfrac{11}{27}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}11\left|x+3\right|=54\\11\left(\left|y\right|-2\right)=27\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left|x+3\right|=\dfrac{54}{11}\\\left|y\right|=\dfrac{27}{11}+2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x+3=\dfrac{54}{11}\\x+3=\dfrac{-54}{11}\end{matrix}\right.\\\left[{}\begin{matrix}y=\dfrac{27}{11}+2\\y=-\left(\dfrac{27}{11}+2\right)\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=\dfrac{21}{11}\left(TM\right)\\x=\dfrac{-87}{11}\left(TM\right)\end{matrix}\right.\\\left[{}\begin{matrix}y=\dfrac{49}{11}\left(TM\right)\\y=-\dfrac{49}{11}\left(TM\right)\end{matrix}\right.\end{matrix}\right.\\ Vậy:\left(x;y\right)=\left\{\left(\dfrac{21}{11};\dfrac{49}{11}\right);\left(\dfrac{-87}{11};\dfrac{49}{11}\right);\left(\dfrac{21}{11};\dfrac{-49}{11}\right);\left(\dfrac{-87}{11};\dfrac{-49}{11}\right)\right\}\)

\(1,\) thiếu đề

\(2,\dfrac{5x+2}{6}-\dfrac{8x-1}{3}=\dfrac{4x+2}{5}-5\)

\(\Leftrightarrow\dfrac{5\left(5x+2\right)}{30}-\dfrac{10\left(8x-1\right)}{30}=\dfrac{6\left(4x+2\right)}{30}-\dfrac{150}{30}\)

\(\Leftrightarrow5\left(5x+2\right)-10\left(8x-1\right)=6\left(4x+2\right)-150\)

\(\Leftrightarrow25x+10-80x+10=24x+12-150\)

\(\Leftrightarrow-55x+20=24x-138\)

\(\Leftrightarrow24x-138+55x-20=0\)

\(\Leftrightarrow79x-158=0\)

\(\Leftrightarrow x=2\)

\(3,ĐKXĐ:\left\{{}\begin{matrix}x\ne1\\x\ne-1\\x\ne3\end{matrix}\right.\\ \dfrac{x}{2x-6}+\dfrac{x}{2x-2}=\dfrac{-2x}{\left(x+1\right)\left(3-x\right)}\)

\(\Leftrightarrow\dfrac{x}{2\left(x-3\right)}+\dfrac{x}{2\left(x-1\right)}+\dfrac{2x}{\left(x+1\right)\left(3-x\right)}=0\)

\(\Leftrightarrow\dfrac{x}{2\left(x-3\right)}+\dfrac{x}{2\left(x-1\right)}-\dfrac{2x}{\left(x+1\right)\left(x-3\right)}=0\)

\(\Leftrightarrow x\left(\dfrac{1}{2\left(x-3\right)}+\dfrac{1}{2\left(x-1\right)}-\dfrac{2}{\left(x+1\right)\left(x-3\right)}\right)=0\)

\(\Leftrightarrow x\left(\dfrac{\left(x-1\right)\left(x+1\right)}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}+\dfrac{\left(x-3\right)\left(x+1\right)}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}-\dfrac{4\left(x-1\right)}{2\left(x+1\right)\left(x-3\right)\left(x-1\right)}\right)=0\)

\(\Leftrightarrow x\left(\dfrac{x^2-1}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}+\dfrac{x^2-2x-3}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}-\dfrac{4x-4}{2\left(x+1\right)\left(x-3\right)\left(x-1\right)}\right)=0\)

\(\Leftrightarrow x.\dfrac{x^2-1+x^2-2x-3-4x+4}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}=0\)

\(\Leftrightarrow x.\dfrac{2x^2-6x}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}=0\)

\(\Leftrightarrow x.\dfrac{2x\left(x-3\right)}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}=0\)

\(\Leftrightarrow x.\dfrac{x}{\left(x-1\right)\left(x+1\right)}=0\)

\(\Leftrightarrow\dfrac{x^2}{\left(x-1\right)\left(x+1\right)}=0\)

\(\Leftrightarrow x=0\)

ĐKXĐ: \(x\notin\left\{-1;-2;-3;-4\right\}\)

Ta có: \(\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\)

\(\Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}=\dfrac{1}{6}\)

\(\Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+4}=\dfrac{1}{6}\)

\(\Leftrightarrow\dfrac{x+4}{\left(x+1\right)\left(x+4\right)}-\dfrac{x+1}{\left(x+1\right)\left(x+4\right)}=\dfrac{1}{6}\)

\(\Leftrightarrow\dfrac{x+4-x-1}{\left(x+1\right)\left(x+4\right)}=\dfrac{x^2+5x+4}{6\left(x+1\right)\left(x+4\right)}\)

\(\Leftrightarrow\dfrac{18}{6\left(x+1\right)\left(x+4\right)}=\dfrac{x^2+5x+4}{6\left(x+1\right)\left(x+4\right)}\)

Suy ra: \(x^2+5x+4=18\)

\(\Leftrightarrow x^2+5x-14=0\)

\(\Leftrightarrow x^2+7x-2x-14=0\)

\(\Leftrightarrow x\left(x+7\right)-2\left(x+7\right)=0\)

\(\Leftrightarrow\left(x+7\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+7=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-7\left(nhận\right)\\x=2\left(nhận\right)\end{matrix}\right.\)

Vậy: S={-7;2}

ĐKXĐ: $x \neq -1;-2;-3;-4$

$pt⇔\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}=\dfrac{1}{6}$

$⇔\dfrac{1}{x+1}-\dfrac{1}{x+4}=\dfrac{1}{6}$

$⇔\dfrac{3}{(x+1)(x+4)}=\dfrac{1}{6}$

$⇔x^2+5x+4=18$

$⇔x^2+5x-14=0$

$⇔(x-2)(x+7)=0$

$⇔$ \(\left[{}\begin{matrix}x=2\\x=-7\end{matrix}\right.\)(t/m)

Vậy...